# Why does Mascolell define second-order stochastic dominance as such?

Is not Mascolell's definition in his microeconomic theory of the second-order stochastic dominance narrower? He defines for distribution functions with the same mean only. Although he gives some motivation for doing this, somehow I do not get his motivation.

The motivation is I believe clearly stated in p. 197 beginning of the section, where they write that they want to use Second-order stochastic dominance to reflect comparisons related to "riskiness/dispersion".

To make this prominent, they use distributions with the same mean, so that it is obvious that we compare them not w.r.t their means, but w.r.t. riskiness/dispersion. In fact, the authors use the concept as equivalent to the stochastic dominance over mean-preserving spreads (see proposition 6.D.2 p. 199).

Using the general definition (which is defined by what they provide as property 6.D.2 p. 198) would not have offered them something more, in view of the reasons they want to use the concept.

• Thank you. A natural question that follows is why they did not do the same thing for first-order stochastic dominance?? – Megadeth Oct 20 '17 at 23:10
• @GaryMoore Because first-order SD is all about expected (mean) returns, in fact it is equivalent to say that the dominant has higher mean that the dominated. – Alecos Papadopoulos Oct 20 '17 at 23:47
• @AlecosPapadopoulos If lottery $F$ fosds lottery $G$, then $E[F] \ge E[G]$. However, it is not true that whenever $E[F] \ge E[G]$, $F$ fosds $G$. For a counter-example, see Wikipedia. Hence, first-order stochastic dominance is more than just about the mean, and it is not equivalent to say that the dominant has a higher mean than the dominated. – Theoretical Economist Oct 21 '17 at 13:33
• @TheoreticalEconomist Thanks for spotting this. Indeed I had in mind expected utility "returns", $E[u(F)] \ge E[u(G)]$ for non-decreasing $u$, but I forgot the "utility" part and resulted in a mistaken assertion. – Alecos Papadopoulos Oct 21 '17 at 16:12